1. **Problem Statement:** A firm manufactures two products with profits of 3 and 5 per unit respectively. Each product requires processing time in two departments D1 and D2. Product 1 requires 1 minute in D1 and 3 minutes in D2. Product 2 requires 2 minutes in D1 and 2 minutes in D2. Available machine time is 860 minutes in D1 and 1200 minutes in D2. We want to find how many units of each product to produce to maximize profit. 2. **Define variables:** Let $x$ = units of product 1 produced per day Let $y$ = units of product 2 produced per day 3. **Objective function (profit to maximize):** $$Z = 3x + 5y$$ 4. **Constraints based on machine time:** - For department D1: $$1x + 2y \leq 860$$ - For department D2: $$3x + 2y \leq 1200$$ - Non-negativity: $$x \geq 0, y \geq 0$$ 5. **Graphical method:** Plot the constraints: - From D1: $$y \leq \frac{860 - x}{2}$$ - From D2: $$y \leq \frac{1200 - 3x}{2}$$ 6. **Find intersection points of constraints and axes:** - Intersection with axes for D1: - When $x=0$, $y=430$ - When $y=0$, $x=860$ - Intersection with axes for D2: - When $x=0$, $y=600$ - When $y=0$, $x=400$ 7. **Find intersection of the two constraint lines:** Solve system: $$\begin{cases} x + 2y = 860 \\ 3x + 2y = 1200 \end{cases}$$ Subtract first from second: $$3x + 2y - (x + 2y) = 1200 - 860 \Rightarrow 2x = 340 \Rightarrow x = 170$$ Substitute $x=170$ into first: $$170 + 2y = 860 \Rightarrow 2y = 690 \Rightarrow y = 345$$ 8. **Evaluate objective function at corner points:** - At $(0,0)$: $Z=0$ - At $(0,430)$: $Z=3(0)+5(430)=2150$ - At $(400,0)$: $Z=3(400)+5(0)=1200$ - At $(170,345)$: $Z=3(170)+5(345)=510+1725=2235$ 9. **Conclusion:** Maximum profit is $2235$ when producing $170$ units of product 1 and $345$ units of product 2 per day.